I’m a PhD student in Genetics/proteonomics. I’m taking this course to refresh my R skills and learn some new stuff.
Data wrangling exersize is in the data folder.
This dataset contains information about students and their attitude/strategy towards learning and their exam scores.
setwd("~/GitHub/IODS-project/data")
data <- read.table("learning2014.csv", header = TRUE, sep = ",")
head(data)
## Gender Age Attitude Points Deep Surf Stra
## 1 F 53 37 25 3.583333 2.583333 3.375
## 2 M 55 31 12 2.916667 3.166667 2.750
## 3 F 49 25 24 3.500000 2.250000 3.625
## 4 M 53 35 10 3.500000 2.250000 3.125
## 5 M 49 37 22 3.666667 2.833333 3.625
## 6 F 38 38 21 4.750000 2.416667 3.625
names(data)
## [1] "Gender" "Age" "Attitude" "Points" "Deep" "Surf"
## [7] "Stra"
dim(data)
## [1] 183 7
There are 7 variables and 183 people.
library(ggplot2)
summary(data)
## Gender Age Attitude Points Deep
## F:122 Min. :17.00 Min. :14.00 Min. : 0.00 Min. :1.583
## M: 61 1st Qu.:21.00 1st Qu.:26.00 1st Qu.:18.00 1st Qu.:3.333
## Median :22.00 Median :32.00 Median :22.00 Median :3.667
## Mean :25.58 Mean :31.21 Mean :20.61 Mean :3.696
## 3rd Qu.:27.00 3rd Qu.:37.00 3rd Qu.:26.00 3rd Qu.:4.083
## Max. :55.00 Max. :50.00 Max. :33.00 Max. :4.917
## Surf Stra
## Min. :1.583 Min. :1.250
## 1st Qu.:2.417 1st Qu.:2.562
## Median :2.833 Median :3.125
## Mean :2.792 Mean :3.085
## 3rd Qu.:3.167 3rd Qu.:3.625
## Max. :4.333 Max. :5.000
Summary of all variables. Dataset is about students (Age, Gender) and their attitudes towards learning (Attitude, Deep, Stra, Surf) and their combined scores (Points). Combination stats like Deep/Surf/Stra combine several questions from the questionairre togetger.
Deep is a combination of questionaire questions that reflect “Deep Approach” to learning. (Seeking Meaning, Relating Ideas, Use of Evidence)
Surf is a “Surface Approach” to learning. (Lack of Purpose, Unrelated Memorizing, Syllabus-boundness)
Stra is a “Strategic Approach” to learning. (Organized Studying, Time Management)
I will use the graphical library ggplot to do graphics.
#Pie chart of gender
ggplot(data, aes(x=factor(1), fill = factor(Gender)))+geom_bar(width=1)+coord_polar(theta="y")+theme_void()+labs(title="Gender")
There are 122 females and 61 males.
#Histogram of Age
qplot(data$Age, geom="histogram")+labs(title="Histogram for Age") +
labs(x="Age", y="Count")
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
Age distribution is slanted, there are a lot more people in their 20’s than other people. Age groups cannot really be compared directly because the amount of people are so different.
#Boxplot of gender and points
ggplot(data, aes(x=Gender, y=Points, fill=Gender)) +
geom_boxplot()+labs(title="Gender and amount of points")
It looks like Males have slightly more points by average than females (black bar in the middle is average), but the difference might not be significant. Also there’s a lot more females (122) than males so that might affect this.
Points and Attitude:
#Linear regression of Attitude and Points
ggplot(data, aes(x = Attitude, y = Points)) + geom_point() + geom_smooth(method="lm")+labs(title="Linear regression of attitude and points")
cor.test(data$Attitude, data$Points, method="pearson")
##
## Pearson's product-moment correlation
##
## data: data$Attitude and data$Points
## t = 4.8513, df = 181, p-value = 2.635e-06
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
## 0.2042082 0.4615619
## sample estimates:
## cor
## 0.3392167
Test of simple linear regression. Pearson correlation is significant and low positive, 0.339. (Pearson correlation assumes normalcy, not sure if it applies here)
Scatter plot matrix:
library(GGally)
p <- ggpairs(data, mapping = aes(col=Gender, alpha=0.3), lower = list(combo = wrap("facethist", bins = 20)))
p
Compilation of all sorts of different statistics.
Target: Points
linear <- lm(Points ~ Gender+Age+Attitude, data=data)
summary(linear)
##
## Call:
## lm(formula = Points ~ Gender + Age + Attitude, data = data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -25.153 -2.520 1.716 5.411 13.022
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 10.23020 3.39239 3.016 0.00294 **
## GenderM -0.39852 1.35699 -0.294 0.76934
## Age -0.08766 0.07944 -1.103 0.27133
## Attitude 0.40862 0.08707 4.693 5.33e-06 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 8.196 on 179 degrees of freedom
## Multiple R-squared: 0.1218, Adjusted R-squared: 0.1071
## F-statistic: 8.278 on 3 and 179 DF, p-value: 3.465e-05
This correlation is not significant for gender and age, but is for attitude. I will keep Attitude and remove Gender, Age.
summary(lm(Points~Attitude+Stra+Surf, data=data))
##
## Call:
## lm(formula = Points ~ Attitude + Stra + Surf, data = data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -25.151 -3.212 2.233 5.257 13.694
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 4.77806 5.15767 0.926 0.3555
## Attitude 0.37570 0.08308 4.522 1.11e-05 ***
## Stra 1.89680 0.78318 2.422 0.0164 *
## Surf -0.62623 1.14945 -0.545 0.5866
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 8.082 on 179 degrees of freedom
## Multiple R-squared: 0.146, Adjusted R-squared: 0.1317
## F-statistic: 10.2 on 3 and 179 DF, p-value: 3.089e-06
Linear regression is significant for Attitude and Stra, but not Surf. Remove Surf.
summary(lm(Points~Attitude+Stra+Deep, data=data))
##
## Call:
## lm(formula = Points ~ Attitude + Stra + Deep, data = data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -23.499 -2.557 1.755 5.097 15.420
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 9.69651 4.84553 2.001 0.04689 *
## Attitude 0.40408 0.08178 4.941 1.78e-06 ***
## Stra 2.07504 0.77411 2.681 0.00804 **
## Deep -2.19240 1.08585 -2.019 0.04497 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 7.998 on 179 degrees of freedom
## Multiple R-squared: 0.1637, Adjusted R-squared: 0.1496
## F-statistic: 11.68 on 3 and 179 DF, p-value: 5.036e-07
Attitude, Strategy, and trying to get a Deep understanding of the subject during studying explains exam results the best. Here multiple linear regression and all subregressions are significant. So best predictors of a good exam score is the overall attitude of the student and study habits that aim at: Seeking Meaning, Relating Ideas, Use of Evidence, Organized Studying, and Time Management.
R-squared is a statistical measure of how close the data are to the fitted regression line. If the data formed linearly on the regression line, R-squared would be 1 or close to 1. Multiple R-squared is low and that says that these variables do not explain the whole variation in exam points. There are other factors that affect the distribution/orientation of the data.
ggplot(data, aes(x=Points, y=Attitude+Stra+Deep))+geom_point()+geom_smooth(method="lm")+labs(title="Linear regression of points and attitude+stra+deep")
In the picture there’s slight positive linear correlation between Points and Attitude+Stra+Deep, meaning higher scores for an individual in Attitude+Stra+Deep generally mean there’s also a higher Points. While Points correlates with Attitude+Stra+Deep, those variables do not explain all variation in Points. Other factors that could affect points might be how familiar the student is with the course material previously and how much absolute time they spent on studying.
linear <- lm(Points~Attitude+Stra+Deep, data=data)
par(mfrow = c(2,2))
plot(linear, which = c(1,2,5))
Q-Q Plot shows that the left tail of the distribution does not follow normalcy. It deviates a lot. I assume left tail contains all the “0 points” people, who answered the questions but possibly didn’t show up for the exam (0 points.) These people should be in a real analysis of the data.
There is also high spread in residuals, but they look mostly grouped linearly.
There is one higher leverage outlier in the data at 0.10 Leverage, while others are mostly between 0.0-0.04.
setwd("~/GitHub/IODS-project/data")
alc <- read.table("~/GitHub/IODS-project/data/alc.csv", header = TRUE, sep = ",")
names(alc)
## [1] "school" "sex" "age" "address" "famsize"
## [6] "Pstatus" "Medu" "Fedu" "Mjob" "Fjob"
## [11] "reason" "nursery" "internet" "guardian" "traveltime"
## [16] "studytime" "failures" "schoolsup" "famsup" "paid"
## [21] "activities" "higher" "romantic" "famrel" "freetime"
## [26] "goout" "Dalc" "Walc" "health" "absences"
## [31] "G1" "G2" "G3" "alc_use" "high_use"
This is a dataset made from two Portuguese school data attributes include descriptors of students (like age, sex, family size, profession of mom and dad, grades etc) and alcohol usage (Dalc = workday alcohol consumption, Walc = weekend alcohol consumption). Alc_use combines Dalc and Walk, and high_use is alcohol usage that’s higher than 2 (1- very low, 5- very high).
Legal drinking age in Portugal is 16 for beer/cider and 18 for hard liquor. I would hope that there are significantly more high alchohol users in students that are of legal drinking age.
library(ggplot2)
alc$legal <- ifelse(alc$age >= 16, TRUE, FALSE)
ggplot(data = alc, aes(x=high_use))+geom_bar()+facet_wrap("legal")+geom_text(stat='count', aes(label=..count..))
#Students under legal drinking age that are high users
18/(63+18)
## [1] 0.2222222
#Students over legal drinking age that are high users
96/(205+96)
## [1] 0.3189369
Here groupings are based on is the student legal drinking age (TRUE/FALSE). The graph shows that there are much more responders that are of legal drinking age. Percentage of students that are high users and under the legal drinking age is 22% and number of students that are high users and over the legal drinking age is 32%. There’s a noticable difference.
26 goout - going out with friends (numeric: from 1 - very low to 5 - very high)
Assumption: Kids that go out alot with their friends are more likely to drink than those who don’t.
ggplot(data = alc, aes(x=high_use))+geom_bar()+facet_wrap("goout")+geom_text(stat='count', aes(label=..count..))
ftable(table(alc$high_use, alc$goout))
## 1 2 3 4 5
##
## FALSE 19 84 103 41 21
## TRUE 3 16 23 40 32
Here it’s very obvious, kids that go out a lot drink a lot more than kids that don’t go out much.
Assumption: Kids who get more than half the points in every grade (First period, second period, final grade) are less likely to drink a lot than kids than don’t even manage half the points.
alc$grades <- ifelse(alc$G1 >= 11 & alc$G2 >= 11 & alc$G3 >=11, TRUE, FALSE)
ggplot(data = alc, aes(x=high_use))+geom_bar()+facet_wrap("grades")+geom_text(stat='count', aes(label=..count..))
#false
66/(66+118)
## [1] 0.3586957
#true
48/(48+150)
## [1] 0.2424242
This looks like it might be significant, kids that get more than half the points are 24% high alcohol users and kids that don’t are 35.9% high alchohol users.
Assumption: Kids who spend more time studying drink less.
ggplot(data = alc, aes(x=high_use))+geom_bar()+facet_wrap("studytime")+geom_text(stat='count', aes(label=..count..))
ftable(table(alc$high_use, alc$studytime))
## 1 2 3 4
##
## FALSE 58 135 52 23
## TRUE 42 60 8 4
#1 < 2hours
42/(42+58)
## [1] 0.42
#2 2 to 5 hours
60/(60+135)
## [1] 0.3076923
#3 5 to 10 hours
8/(8+52)
## [1] 0.1333333
#4 over 10 hours
4/(4+23)
## [1] 0.1481481
Here it’s pretty clear, kids that study more than 5 hours a week are much less likely to be drinkers (14.8%, 13.3%) and kids who study less are much more likely to be heavy drinkers ( 30.7% and 42%). Especially if the kids spend almost no time studying (<2).
m <- glm(high_use ~ legal, data = alc, family = "binomial")
summary(m)
##
## Call:
## glm(formula = high_use ~ legal, family = "binomial", data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -0.8765 -0.8765 -0.8765 1.5118 1.7344
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.2528 0.2673 -4.687 2.77e-06 ***
## legalTRUE 0.4941 0.2945 1.678 0.0934 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 465.68 on 381 degrees of freedom
## Residual deviance: 462.70 on 380 degrees of freedom
## AIC: 466.7
##
## Number of Fisher Scoring iterations: 4
coef(m)
## (Intercept) legalTRUE
## -1.2527630 0.4941012
confint(m)
## Waiting for profiling to be done...
## 2.5 % 97.5 %
## (Intercept) -1.80548222 -0.7512117
## legalTRUE -0.06548363 1.0947388
Age is not significant factor in 95% statistical significance, but would be 90% statistical significance.
m <- glm(high_use ~ goout, data = alc, family = "binomial")
summary(m)
##
## Call:
## glm(formula = high_use ~ goout, family = "binomial", data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.3721 -0.7672 -0.5450 0.9945 2.3082
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -3.3512 0.4151 -8.073 6.84e-16 ***
## goout 0.7596 0.1157 6.564 5.23e-11 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 465.68 on 381 degrees of freedom
## Residual deviance: 415.68 on 380 degrees of freedom
## AIC: 419.68
##
## Number of Fisher Scoring iterations: 4
coef(m)
## (Intercept) goout
## -3.3511801 0.7596076
confint(m)
## Waiting for profiling to be done...
## 2.5 % 97.5 %
## (Intercept) -4.1958512 -2.5655359
## goout 0.5385448 0.9931232
How often kids go “out” with their friends is significant factor that explains high alcohol use.
m <- glm(high_use ~ grades, data = alc, family = "binomial")
summary(m)
##
## Call:
## glm(formula = high_use ~ grades, family = "binomial", data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -0.9426 -0.9426 -0.7452 1.4320 1.6835
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -0.5810 0.1537 -3.78 0.000157 ***
## gradesTRUE -0.5584 0.2261 -2.47 0.013526 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 465.68 on 381 degrees of freedom
## Residual deviance: 459.51 on 380 degrees of freedom
## AIC: 463.51
##
## Number of Fisher Scoring iterations: 4
coef(m)
## (Intercept) gradesTRUE
## -0.5810299 -0.5584044
confint(m)
## Waiting for profiling to be done...
## 2.5 % 97.5 %
## (Intercept) -0.8871701 -0.2834717
## gradesTRUE -1.0051883 -0.1174594
Bad grades mean kid is more likely to be a high alcohol user and higher grades mean low chance.
m <- glm(high_use ~ studytime, data = alc, family = "binomial")
summary(m)
##
## Call:
## glm(formula = high_use ~ studytime, family = "binomial", data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.0603 -0.8314 -0.8314 1.2993 2.1010
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.3209 0.3076 1.043 0.297
## studytime -0.6029 0.1530 -3.941 8.1e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 465.68 on 381 degrees of freedom
## Residual deviance: 448.31 on 380 degrees of freedom
## AIC: 452.31
##
## Number of Fisher Scoring iterations: 4
coef(m)
## (Intercept) studytime
## 0.3209036 -0.6028561
confint(m)
## Waiting for profiling to be done...
## 2.5 % 97.5 %
## (Intercept) -0.2772274 0.9308876
## studytime -0.9123998 -0.3115583
If a kid spends a lot tipe studying they are less likely to be high alchohol users.
Go out, Grades, STudytime
library(dplyr)
##
## Attaching package: 'dplyr'
## The following object is masked from 'package:GGally':
##
## nasa
## The following objects are masked from 'package:stats':
##
## filter, lag
## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, union
m <- glm(high_use ~ studytime+grades+goout, data = alc, family = "binomial")
summary(m)
##
## Call:
## glm(formula = high_use ~ studytime + grades + goout, family = "binomial",
## data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.6683 -0.8015 -0.5478 0.9613 2.5753
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.9907 0.5184 -3.840 0.000123 ***
## studytime -0.5737 0.1659 -3.459 0.000543 ***
## gradesTRUE -0.3014 0.2485 -1.213 0.225280
## goout 0.7340 0.1171 6.269 3.64e-10 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 465.68 on 381 degrees of freedom
## Residual deviance: 399.34 on 378 degrees of freedom
## AIC: 407.34
##
## Number of Fisher Scoring iterations: 4
coef(m)
## (Intercept) studytime gradesTRUE goout
## -1.9906826 -0.5736953 -0.3013601 0.7340343
confint(m)
## Waiting for profiling to be done...
## 2.5 % 97.5 %
## (Intercept) -3.0311038 -0.9938476
## studytime -0.9097114 -0.2575669
## gradesTRUE -0.7897173 0.1864329
## goout 0.5103598 0.9703992
p <- predict(m, type="response")
# add the predicted probabilities to 'alc'
alc <- mutate(alc, probability = p)
# use the probabilities to make a prediction of high_use
alc <- mutate(alc, prediction = probability > 0.5)
table(high_use = alc$high_use, prediction = alc$prediction) %>% prop.table() %>% addmargins()
## prediction
## high_use FALSE TRUE Sum
## FALSE 0.64659686 0.05497382 0.70157068
## TRUE 0.19109948 0.10732984 0.29842932
## Sum 0.83769634 0.16230366 1.00000000
g <- ggplot(alc, aes(x = high_use, y = probability, col=prediction))
g + geom_point()
# define a loss function (mean prediction error)
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
# call loss_func to compute the average number of wrong predictions in the (training) data
loss_func(class = alc$high_use, prob = alc$probability)
## [1] 0.2460733
There’s a 25% chance of a wrong prediction, so I don’t think my variables are a excellent predictor. It would predict “mostly”, but there’s such a big chance of error it’s not good/usable. Also there’s probably a lot of “overlap” between go out and studytime and studytime and grades… Different predictors would have maybe captured more of the actual problem?
library(boot)
cv <- cv.glm(data = alc, cost = loss_func, glmfit = m, K = 10)
# average number of wrong predictions in the cross validation
cv$delta[1]
## [1] 0.2513089
Datacamp model was 0.26 and mine is 0.25… SO they are pretty much the same?
Trying to get a better model:
m <- glm(high_use ~ studytime+failures+goout+absences+sex, data = alc, family = "binomial")
summary(m)
##
## Call:
## glm(formula = high_use ~ studytime + failures + goout + absences +
## sex, family = "binomial", data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.0668 -0.7736 -0.5080 0.7663 2.4984
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -3.27654 0.59805 -5.479 4.28e-08 ***
## studytime -0.38032 0.17414 -2.184 0.028966 *
## failures 0.24195 0.21713 1.114 0.265141
## goout 0.71219 0.12037 5.917 3.28e-09 ***
## absences 0.07748 0.02242 3.456 0.000548 ***
## sexM 0.76991 0.26620 2.892 0.003825 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 465.68 on 381 degrees of freedom
## Residual deviance: 380.06 on 376 degrees of freedom
## AIC: 392.06
##
## Number of Fisher Scoring iterations: 4
coef(m)
## (Intercept) studytime failures goout absences sexM
## -3.27653834 -0.38032241 0.24195066 0.71218501 0.07747656 0.76991496
confint(m)
## Waiting for profiling to be done...
## 2.5 % 97.5 %
## (Intercept) -4.48345712 -2.13293535
## studytime -0.73128369 -0.04615332
## failures -0.18407800 0.67190587
## goout 0.48205204 0.95506124
## absences 0.03450819 0.12366371
## sexM 0.25194047 1.29791469
p <- predict(m, type="response")
# add the predicted probabilities to 'alc'
alc <- mutate(alc, probability = p)
# use the probabilities to make a prediction of high_use
alc <- mutate(alc, prediction = probability > 0.5)
table(high_use = alc$high_use, prediction = alc$prediction) %>% prop.table() %>% addmargins()
## prediction
## high_use FALSE TRUE Sum
## FALSE 0.65968586 0.04188482 0.70157068
## TRUE 0.17015707 0.12827225 0.29842932
## Sum 0.82984293 0.17015707 1.00000000
g <- ggplot(alc, aes(x = high_use, y = probability, col=prediction))
g + geom_point()
# define a loss function (mean prediction error)
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
# call loss_func to compute the average number of wrong predictions in the (training) data
loss_func(class = alc$high_use, prob = alc$probability)
## [1] 0.2120419
cv <- cv.glm(data = alc, cost = loss_func, glmfit = m, K = 10)
# average number of wrong predictions in the cross validation
cv$delta[1]
## [1] 0.2251309
With studytime+failures+goout+absences+sex it drops to 22.5… So maybe that’s good enough!
data("Boston")
# explore the dataset
str(Boston)
## 'data.frame': 506 obs. of 14 variables:
## $ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
## $ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
## $ rm : num 6.58 6.42 7.18 7 7.15 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
## $ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $ rad : int 1 2 2 3 3 3 5 5 5 5 ...
## $ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
## $ black : num 397 397 393 395 397 ...
## $ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
## $ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...
dim(Boston)
## [1] 506 14
summary(Boston)
## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08204 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00
# calculate the correlation matrix and round it
cor_matrix<-cor(Boston) %>% round(digits = 2)
# print the correlation matrix
cor_matrix
## crim zn indus chas nox rm age dis rad tax
## crim 1.00 -0.20 0.41 -0.06 0.42 -0.22 0.35 -0.38 0.63 0.58
## zn -0.20 1.00 -0.53 -0.04 -0.52 0.31 -0.57 0.66 -0.31 -0.31
## indus 0.41 -0.53 1.00 0.06 0.76 -0.39 0.64 -0.71 0.60 0.72
## chas -0.06 -0.04 0.06 1.00 0.09 0.09 0.09 -0.10 -0.01 -0.04
## nox 0.42 -0.52 0.76 0.09 1.00 -0.30 0.73 -0.77 0.61 0.67
## rm -0.22 0.31 -0.39 0.09 -0.30 1.00 -0.24 0.21 -0.21 -0.29
## age 0.35 -0.57 0.64 0.09 0.73 -0.24 1.00 -0.75 0.46 0.51
## dis -0.38 0.66 -0.71 -0.10 -0.77 0.21 -0.75 1.00 -0.49 -0.53
## rad 0.63 -0.31 0.60 -0.01 0.61 -0.21 0.46 -0.49 1.00 0.91
## tax 0.58 -0.31 0.72 -0.04 0.67 -0.29 0.51 -0.53 0.91 1.00
## ptratio 0.29 -0.39 0.38 -0.12 0.19 -0.36 0.26 -0.23 0.46 0.46
## black -0.39 0.18 -0.36 0.05 -0.38 0.13 -0.27 0.29 -0.44 -0.44
## lstat 0.46 -0.41 0.60 -0.05 0.59 -0.61 0.60 -0.50 0.49 0.54
## medv -0.39 0.36 -0.48 0.18 -0.43 0.70 -0.38 0.25 -0.38 -0.47
## ptratio black lstat medv
## crim 0.29 -0.39 0.46 -0.39
## zn -0.39 0.18 -0.41 0.36
## indus 0.38 -0.36 0.60 -0.48
## chas -0.12 0.05 -0.05 0.18
## nox 0.19 -0.38 0.59 -0.43
## rm -0.36 0.13 -0.61 0.70
## age 0.26 -0.27 0.60 -0.38
## dis -0.23 0.29 -0.50 0.25
## rad 0.46 -0.44 0.49 -0.38
## tax 0.46 -0.44 0.54 -0.47
## ptratio 1.00 -0.18 0.37 -0.51
## black -0.18 1.00 -0.37 0.33
## lstat 0.37 -0.37 1.00 -0.74
## medv -0.51 0.33 -0.74 1.00
# visualize the correlation matrix
corrplot(cor_matrix, method="circle", type="upper", cl.pos="b", tl.pos="d", tl.cex = 0.6)
Boston Dataset describes Housing Values in Suburbs of Boston. Here positive correlations are displayed in blue and negative correlations in red color. Color intensity and the size of the circle are proportional to the correlation coefficients.
f.ex. Age is strongly negatively correlated with weighted mean of distances to five Boston employment centres (dis). And index of accessibility to radial highways (rad) is strongly positively correlated with property tax. Number of rooms is positively corrrelated with median value of homes, whereas lower status of the population is strongly negatively correlated.
# center and standardize variables
boston_scaled <- scale(Boston)
# summaries of the scaled variables
summary(boston_scaled)
## crim zn indus
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668
## Median :-0.390280 Median :-0.48724 Median :-0.2109
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202
## chas nox rm age
## Min. :-0.2723 Min. :-1.4644 Min. :-3.8764 Min. :-2.3331
## 1st Qu.:-0.2723 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366
## Median :-0.2723 Median :-0.1441 Median :-0.1084 Median : 0.3171
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.:-0.2723 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059
## Max. : 3.6648 Max. : 2.7296 Max. : 3.5515 Max. : 1.1164
## dis rad tax ptratio
## Min. :-1.2658 Min. :-0.9819 Min. :-1.3127 Min. :-2.7047
## 1st Qu.:-0.8049 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876
## Median :-0.2790 Median :-0.5225 Median :-0.4642 Median : 0.2746
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6617 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058
## Max. : 3.9566 Max. : 1.6596 Max. : 1.7964 Max. : 1.6372
## black lstat medv
## Min. :-3.9033 Min. :-1.5296 Min. :-1.9063
## 1st Qu.: 0.2049 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median : 0.3808 Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.4332 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 0.4406 Max. : 3.5453 Max. : 2.9865
# class of the boston_scaled object
class(boston_scaled)
## [1] "matrix"
# change the object to data frame
boston_scaled <- as.data.frame(boston_scaled)
# summary of the scaled crime rate
summary(boston_scaled$crim)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -0.419367 -0.410563 -0.390280 0.000000 0.007389 9.924110
# create a quantile vector of crim and print it
bins <- quantile(boston_scaled$crim)
bins
## 0% 25% 50% 75% 100%
## -0.419366929 -0.410563278 -0.390280295 0.007389247 9.924109610
# create a categorical variable 'crime'
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, labels = c("low", "med_low", "med_high", "high"))
# look at the table of the new factor crime
table(crime)
## crime
## low med_low med_high high
## 127 126 126 127
# remove original crim from the dataset
boston_scaled <- dplyr::select(boston_scaled, -crim)
# add the new categorical value to scaled data
boston_scaled <- data.frame(boston_scaled, crime)
# number of rows in the Boston dataset
n <- nrow(boston_scaled)
# choose randomly 80% of the rows
ind <- sample(n, size = n * 0.8)
# create train set
train <- boston_scaled[ind,]
# create test set
test <- boston_scaled[-ind,]
# save the correct classes from test data
correct_classes <- test$crime
# remove the crime variable from test data
test <- dplyr::select(test, -crime)
In the scaling we subtract the column means from the corresponding columns and divide the difference with standard deviation so the values become z-scores. This normalizes the data so now it will be normally distributed. For some multivariate techniques such as multidimensional scaling and cluster analysis, the concept of distance between the units in the data is often of considerable interest and importance. When the variables in a multivariate data set are on different scales, it makes more sense to calculate the distances after some form of standardization.
# linear discriminant analysis
lda.fit <- lda(crime ~ ., data = train)
# print the lda.fit object
lda.fit
## Call:
## lda(crime ~ ., data = train)
##
## Prior probabilities of groups:
## low med_low med_high high
## 0.259901 0.250000 0.240099 0.250000
##
## Group means:
## zn indus chas nox rm
## low 1.04674041 -0.9249028 -0.1223442968 -0.8740729 0.44740737
## med_low -0.09731296 -0.3236341 0.0005392655 -0.6043588 -0.07470053
## med_high -0.37054365 0.2047360 0.0929688923 0.4047620 0.18904966
## high -0.48724019 1.0171306 -0.0774231154 1.0294057 -0.40041345
## age dis rad tax ptratio
## low -0.9010817 0.9163203 -0.6821760 -0.7256223 -0.41497152
## med_low -0.4063844 0.3886193 -0.5509118 -0.5179664 -0.05309119
## med_high 0.4191027 -0.3916045 -0.4052697 -0.3030324 -0.38469937
## high 0.8030977 -0.8496991 1.6379981 1.5139626 0.78062517
## black lstat medv
## low 0.38147837 -0.77250305 0.53977170
## med_low 0.31941250 -0.19990287 0.03680298
## med_high 0.09061597 -0.01694445 0.24561495
## high -0.77839577 0.91309970 -0.68921118
##
## Coefficients of linear discriminants:
## LD1 LD2 LD3
## zn 0.08069079 0.735376091 -0.94059732
## indus 0.03225585 -0.342607958 0.28000290
## chas -0.08537396 -0.004390383 0.16962190
## nox 0.38737717 -0.677037621 -1.32248675
## rm -0.08837978 -0.142888669 -0.09900877
## age 0.22178887 -0.297763827 -0.15052175
## dis -0.06380347 -0.330256124 0.15823936
## rad 3.09834362 0.953981811 0.06783711
## tax 0.08818910 -0.075265917 0.32859053
## ptratio 0.10964107 0.099239020 -0.19897887
## black -0.12833687 -0.003542823 0.08394312
## lstat 0.25786935 -0.249971124 0.30800226
## medv 0.21743981 -0.392302137 -0.23726883
##
## Proportion of trace:
## LD1 LD2 LD3
## 0.9467 0.0394 0.0139
# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "orange", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
# target classes as numeric
classes <- as.numeric(train$crime)
# plot the lda results
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 1)
Here Linear Discriminant 1 explains most of the between group variance.
rad: index of accessibility to radial highways is the most significant factor that predicts higher crime rate.
# predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)
# cross tabulate the results
table(correct = correct_classes, predicted = lda.pred$class)
## predicted
## correct low med_low med_high high
## low 11 11 0 0
## med_low 3 16 6 0
## med_high 0 15 13 1
## high 0 0 0 26
This prediction is very good at predicting high crime rates, but worse at predicting med_low or med_high correctly. ’ ##Task 7
data("Boston")
new_boston <- scale(Boston)
# k-means clustering
km <-kmeans(new_boston, centers = 4)
# plot the Boston dataset with clusters
pairs(new_boston, col = km$cluster)
#K-means might produce different results every time, because it randomly assigns the initial cluster centers. The function set.seed() can be used to deal with that.
set.seed(123)
# determine the number of clusters
k_max <- 10
# calculate the total within sum of squares
twcss <- sapply(1:k_max, function(k){kmeans(new_boston, k)$tot.withinss})
# visualize the results
qplot(x = 1:k_max, y = twcss, geom = 'line')
# k-means clustering
km <-kmeans(new_boston, centers = 2)
# plot the Boston dataset with clusters
pairs(new_boston, col = km$cluster)
One way to check optimal amount of clusters is to look at the total within cluster sum of squares (twcss). When twcss drops a lot the optimal number of clusters is found. Here twcss drops around 2. In the LDA there were two “main clusters” one with high crime and some points from med_high and the other larger cluster was the rest of the data points. This k-means clustering on this data also works best with 2 clusters.
km <-kmeans(new_boston, centers = 3)
# linear discriminant analysis
lda.fit <- lda(crime ~ ., data = train)
# print the lda.fit object
lda.fit
## Call:
## lda(crime ~ ., data = train)
##
## Prior probabilities of groups:
## low med_low med_high high
## 0.259901 0.250000 0.240099 0.250000
##
## Group means:
## zn indus chas nox rm
## low 1.04674041 -0.9249028 -0.1223442968 -0.8740729 0.44740737
## med_low -0.09731296 -0.3236341 0.0005392655 -0.6043588 -0.07470053
## med_high -0.37054365 0.2047360 0.0929688923 0.4047620 0.18904966
## high -0.48724019 1.0171306 -0.0774231154 1.0294057 -0.40041345
## age dis rad tax ptratio
## low -0.9010817 0.9163203 -0.6821760 -0.7256223 -0.41497152
## med_low -0.4063844 0.3886193 -0.5509118 -0.5179664 -0.05309119
## med_high 0.4191027 -0.3916045 -0.4052697 -0.3030324 -0.38469937
## high 0.8030977 -0.8496991 1.6379981 1.5139626 0.78062517
## black lstat medv
## low 0.38147837 -0.77250305 0.53977170
## med_low 0.31941250 -0.19990287 0.03680298
## med_high 0.09061597 -0.01694445 0.24561495
## high -0.77839577 0.91309970 -0.68921118
##
## Coefficients of linear discriminants:
## LD1 LD2 LD3
## zn 0.08069079 0.735376091 -0.94059732
## indus 0.03225585 -0.342607958 0.28000290
## chas -0.08537396 -0.004390383 0.16962190
## nox 0.38737717 -0.677037621 -1.32248675
## rm -0.08837978 -0.142888669 -0.09900877
## age 0.22178887 -0.297763827 -0.15052175
## dis -0.06380347 -0.330256124 0.15823936
## rad 3.09834362 0.953981811 0.06783711
## tax 0.08818910 -0.075265917 0.32859053
## ptratio 0.10964107 0.099239020 -0.19897887
## black -0.12833687 -0.003542823 0.08394312
## lstat 0.25786935 -0.249971124 0.30800226
## medv 0.21743981 -0.392302137 -0.23726883
##
## Proportion of trace:
## LD1 LD2 LD3
## 0.9467 0.0394 0.0139
# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "orange", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
# target classes as numeric
classes <- as.numeric(km$cluster)
# plot the lda results
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 1)
rad is the most influencial linear separator for the clusters. It looks like the k-means clustering with 4 clusters makes clusters where low, med_low, med_high, high etc. mix a lot so the clustering is not perfect.
library(plotly)
##
## Attaching package: 'plotly'
## The following object is masked from 'package:MASS':
##
## select
## The following object is masked from 'package:ggplot2':
##
## last_plot
## The following object is masked from 'package:stats':
##
## filter
## The following object is masked from 'package:graphics':
##
## layout
model_predictors <- dplyr::select(train, -crime)
# check the dimensions
dim(model_predictors)
## [1] 404 13
dim(lda.fit$scaling)
## [1] 13 3
# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color= ~train$crime)
#plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color=~classes)
Can’t get the color with km$cluster to work…